Square root spiral

Correct answer: A. \(\sqrt{n+49}\)

Explanation: The governing concept is the Pythagorean theorem, used in the square-root spiral. If one leg of a right triangle is \(\sqrt{n}\) and the perpendicular leg is 7 units, then the square of the hypotenuse equals the sum of the squares of the two perpendicular legs: \(h^2=(\sqrt n)^2+7^2=n+49\). Taking the positive square root, because a length is positive, gives \(h=\sqrt{n+49}\). Thus option A is correct. Option B adds 7 instead of its square, C multiplies the terms incorrectly, and D uses twice 7 rather than \(7^2\).

Correct answer: D. \(\sqrt{1225}\), exactly at (35)

Explanation: The square root spiral uses the Pythagorean theorem to create the next length. When a perpendicular of 1 unit is drawn on a hypotenuse of length \(\sqrt{1224}\), the old hypotenuse and the new perpendicular become the two perpendicular sides of a right triangle. Their squared lengths must be added.

The new hypotenuse is \(\sqrt{(\sqrt{1224})^2+1^2}=\sqrt{1224+1}=\sqrt{1225}\). Since \(35^2=1225\), this length is exactly 35, not merely a value between 34 and 35. Thus option D is correct. Options A, B and C use the wrong arithmetic or give a nonmatching interval; in particular, \(\sqrt{1223}\) is less than 35.

Correct answer: B. \(\sqrt{528}\) lies between 22 and 23, whereas \(\sqrt{530}\) lies between 23 and 24

Explanation: Since \(22^2=484\) and \(23^2=529\), and \(484<528<529\), we get \(22<\sqrt{528}<23\). Similarly, \(530\) lies between \(23^2=529\) and \(24^2=576\), so \(23<\sqrt{530}<24\). Hence, option B is correct. Option C may seem close, but \(528<529\) means that \(\sqrt{528}\) is less than 23. Exam tip: locate a square root by comparing the number with consecutive perfect squares.

Correct answer: C. \(39\)

Explanation: In a square root spiral, the hypotenuse after \(\sqrt{1520}\) is \(\sqrt{1521}\). Since \(1521=39^2\), we get \(\sqrt{1521}=39\). \(38\) is incorrect because \(38^2=1444\), while \(40^2=1600\). Exam tip: To check whether a square root is an integer, compare the number with nearby perfect squares.

Correct answer: C. √389 and 1 unit

Explanation: The standard square root spiral adds a perpendicular segment of unit length to the preceding hypotenuse. If the preceding hypotenuse is √n, then the Pythagorean theorem gives the next length as √((√n)²+1²)=√(n+1). To construct √390, we must have n+1=390, so n=389. Consequently, the correct preceding hypotenuse is √389 and the new perpendicular is 1 unit, which is option C. Option A uses a 2-unit perpendicular and would give 388+2²=392. Option B treats the target as though it were the preceding hypotenuse, while option D would produce √392 because 391+1=392. Only option C satisfies the exact construction rule and reaches √390.

Correct answer: A. \(\sqrt{1296}=36\)

Explanation: In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on, with the number under the root increasing by 1 each time. Therefore, the hypotenuse after \(\sqrt{1295}\) is \(\sqrt{1296}\). Since \(1296=36^2\), \(\sqrt{1296}=36\). \(\sqrt{1297}\) comes one step later, not immediately after \(\sqrt{1295}\). Exam tip: increase the radicand by 1 first, then check whether it is a perfect square.

Correct answer: B. \(25^2<675<26^2\)

Explanation: \(25^2=625\) and \(26^2=676\). Since \(625<675<676\), the correct inequality is \(25^2<675<26^2\), so \(\sqrt{675}\) lies between 25 and 26. Option A is incorrect because 675 is greater than \(25^2=625\). Exam tip: To locate a square root, compare the number with consecutive perfect squares.

Correct answer: B. 2703

Explanation: In a square root spiral, the hypotenuse after \(\sqrt{n}\) is \(\sqrt{n+1}\). Since this hypotenuse is \(52\), \(\sqrt{n+1}=52\). Squaring gives \(n+1=52^2=2704\), so \(n=2703\). Taking \(2704\) would make the next hypotenuse \(\sqrt{2705}\), so it is not correct. Exam tip: square the given hypotenuse first, then subtract 1 because the question refers to the next term.

Correct answer: C. \(\sqrt{18}\rightarrow\sqrt{19}\rightarrow\sqrt{20}\rightarrow\sqrt{21}\rightarrow\sqrt{22}\)

Explanation: The direct answer is C. In the standard square-root spiral, the radicand increases one unit at each construction step. Starting from sqrt{18}, the successive hypotenuses are sqrt{19}, sqrt{20}, sqrt{21} and sqrt{22}. Thus option C gives the complete increasing order. Option A skips sqrt{19} and sqrt{21}, so it is not the usual complete construction. Option B jumps from sqrt{18} to sqrt{22} and then moves backward, which is not the normal order. Option C is correct because every intermediate step is included. Option D begins at the largest value and decreases, reversing the construction and omitting sqrt{18} and sqrt{19}. Memory cue: add 1 inside each successive square root.

Correct answer: B. \(\sqrt{440}\) lies between \(20\) and \(21\), and \(\sqrt{442}\) lies between \(21\) and \(22\)

Explanation: Since \(20^2=400\) and \(21^2=441\), and \(400<440<441\), we get \(20<\sqrt{440}<21\). Similarly, \(21^2=441\) and \(22^2=484\), and \(441<442<484\), so \(21<\sqrt{442}<22\). Therefore, option B is correct. Options C and D incorrectly place both square roots in the same interval. Exam tip: locate a square root by comparing the number with the nearest perfect squares.

Correct answer: A. The next hypotenuse is (\sqrt{625}=25), and (\sqrt{626}) lies between (25) and (26)

Explanation: After (\sqrt{624}), (\sqrt{625}=25) is formed. Since (25^2<626<26^2), (\sqrt{626}) lies between (25) and (26).

Correct answer: C. \(\sqrt{2499}\) lies between \(49\) and \(50\), and \(\sqrt{2500}=50\)

Explanation: Since \(49^2=2401\) is less than \(2499\), while \(50^2=2500\) is greater than \(2499\), we get \(49<\sqrt{2499}<50\). On the other hand, \(2500=50^2\), so \(\sqrt{2500}=50\). Hence, option C is correct. Option A incorrectly states that \(\sqrt{2499}=50\); it is slightly less than \(50\). Exam tip: compare a number with nearby perfect squares to locate its square root.

Correct answer: B. \(48<\sqrt{2400}<49\)

Explanation: \(48^2=2304\) and \(49^2=2401\). Since \(2304<2400<2401\), taking positive square roots gives \(48<\sqrt{2400}<49\). It is less than \(49\) because \(2400<49^2\). Exam tip: To locate a square root, compare the number with consecutive perfect squares.

Correct answer: B. \(59\)

Explanation: In a square root spiral, the hypotenuse after \(\sqrt{3480}\) is \(\sqrt{3481}\). Since \(3481=59^2\), \(\sqrt{3481}=59\). Also, \(58^2=3364\) and \(60^2=3600\), so neither is correct. Exam tip: check the squares of nearby integers to identify a perfect square quickly.

Correct answer: A. \(\sqrt{143}\) lies between 11 and 12, and \(\sqrt{170}\) lies between 13 and 14.

Explanation: Since \(11^2=121\) and \(12^2=144\), we get \(121<143<144\), so \(11<\sqrt{143}<12\). Similarly, \(13^2=169\) and \(14^2=196\) give \(169<170<196\), so \(13<\sqrt{170}<14\). Hence, option A is correct. Option D incorrectly places \(\sqrt{170}\) between 12 and 13. Exam tip: To locate a square root, compare the number with the nearest perfect squares.

Correct answer: D. ((\sqrt{84})^2+1^2=85)

Explanation: A square root spiral adds one new perpendicular side of length 1 at each stage. Starting with a hypotenuse \(\sqrt{84}\), the next right triangle has side lengths \(\sqrt{84}\) and 1. The new hypotenuse must be calculated from the squares of these lengths, as required by the Pythagorean theorem.

The calculation is \((\sqrt{84})^2+1^2=84+1=85\). Taking the positive square root gives the new hypotenuse \(\sqrt{85}\). Therefore option D is correct. The statement \(\sqrt{84}+1=\sqrt{85}\) in option A is not valid, because the square root of a sum is generally not the sum of square roots or lengths. Multiplication and adding a squared length of 2 are also incorrect.

Correct answer: C. \(64<\sqrt{4224}<65\)

Explanation: We have \(64^2=4096\) and \(65^2=4225\). Since \(4096<4224<4225\), taking positive square roots gives \(64<\sqrt{4224}<65\). Although 4224 is very close to \(4225\), it is still less than \(65^2\), so its square root is less than 65. Exam tip: To locate a square root, compare the number with consecutive perfect squares.

Correct answer: A. The new hypotenuse is \(\sqrt{1936}\) and its value is \(44\)

Explanation: In a square root spiral, each new right triangle is formed by adding a unit side to the previous hypotenuse. Hence, if the previous hypotenuse is \(\sqrt{1935}\), the square of the new hypotenuse is \(1935+1=1936\). Therefore, the new hypotenuse is \(\sqrt{1936}\). Since \(44^2=1936\), its value is \(44\). Option B has the correct radicand but an incorrect value. Exam tip: first add \(1\) to the radicand, then check whether the result is a perfect square.

Correct answer: C. \(59<\sqrt{3599}<60\)

Explanation: We have \(59^2=3481\) and \(60^2=3600\). Since \(3481<3599<3600\), it follows that \(59<\sqrt{3599}<60\). Therefore, its position on the square root spiral is between 59 and 60. The close distractor \(60<\sqrt{3599}<61\) is incorrect because 3599 is less than \(60^2\). Exam tip: locate a square root by comparing the number with consecutive perfect squares.

Correct answer: A. \(\sqrt{1520}\) lies between 38 and 39, while \(\sqrt{1522}\) lies between 39 and 40.

Explanation: We have \(38^2=1444\), \(39^2=1521\), and \(40^2=1600\). Since \(1444<1520<1521\), \(38<\sqrt{1520}<39\). Similarly, \(1521<1522<1600\), so \(39<\sqrt{1522}<40\). Therefore, option A is correct. Options B and C incorrectly place both square roots in the same interval. Exam tip: To locate a square root, compare the number with consecutive perfect squares.

Correct answer: B. \(\sqrt{1369}\)

Explanation: The spiral moves from one square-root length to the next by adding a perpendicular side of length 1. If the old hypotenuse is
sqrt{n}
, then the new hypotenuse has square length
n+1
. This follows because
(
sqrt{n}
)^2+1^2=n+1
. The wording identifies
sqrt{1368}
as the previous hypotenuse, so here
n=1368
.

Substituting this value gives the new length
sqrt{1368+1}=
sqrt{1369}
. Hence option B is correct. The answer is not
sqrt{1367}
, which would be a preceding value, and it is not the unchanged
sqrt{1368}
. The value
sqrt{2736}
would incorrectly double the radicand rather than add one. No decimal approximation is needed.

Correct answer: D. 37

Explanation: In a square root spiral, each new hypotenuse represents the square root of the corresponding number. Since \(1369=37\times37=37^2\), \(\sqrt{1369}=37\). Option 36 is incorrect because \(36^2=1296\), while \(38^2=1444\). Exam tip: check nearby perfect squares to identify a square root quickly.

Correct answer: C. \(\sqrt{3025}\)

Explanation: In a square root spiral, when one leg is \(\sqrt{n}\) and the perpendicular leg is 1 unit, the new hypotenuse is \(\sqrt{n+1}\). Therefore, drawing a 1-unit perpendicular on \(\sqrt{3024}\) gives \(\sqrt{3024+1}=\sqrt{3025}\). \(\sqrt{3023}\) represents the previous stage, while \(\sqrt{3026}\) would be the next stage. Exam tip: add exactly 1 to the radicand at each new step.

Correct answer: A. \(50<\sqrt{2600}<51\)

Explanation: \(50^2=2500\) and \(51^2=2601\). Since \(2500<2600<2601\), taking square roots gives \(50<\sqrt{2600}<51\). \(\sqrt{2600}=51\) is incorrect because \(51^2=2601\), not 2600. Exam tip: To find the interval of a square root, compare the number with the nearest perfect squares.

Correct answer: A. Take the (\sqrt{5}) hypotenuse length in a compass and draw an arc from the origin

Explanation: The hypotenuse length of the square root to be marked is taken in the compass. An arc from the origin gives the correct location.